Connected groups of rank zero are unipotent

If $G$ is unipotent, then $G_{k^{\mathrm{a}}}$ is unipotent and contains no nontrivial torus by [F1]. Conversely, suppose there is no such torus. By field-extension descent of unipotence we may work over $k^{\mathrm{a}}$, and hence assume $k$ algebraically closed. Choose a Borel subgroup $B$ and write $B=B_u⋊T$ by [F2]. Since $T\subseteq G$ and there are no nontrivial tori, $T=1$, so $B=B_u$ is unipotent.

By [F3] choose a representation $W$ and a line $L=kv$ with $B={\mathrm{Stab}}_G(L)$. The one-dimensional representation $L$ of the unipotent group $B$ is trivial by its fixed-vector criterion, so every element of $B$ fixes $v$ scheme-theoretically. Consequently $B={\mathrm{Stab}}_G(v)$: the vector stabilizer lies in the line stabilizer, and the reverse inclusion was just proved. The orbit morphism $g\mapsto gv$ is right $B$-invariant and descends by the fppf quotient to a morphism $f:G\mathrm{/}B\to W$.

The scheme $G\mathrm{/}B$ is complete by [F2], connected as the surjective image of connected $G$, and reduced by [F3]. Thus $f$ has a single closed point as image. It contains $v=f(eB)$, so the point is $v$. Since $G\mathrm{/}B$ is reduced, every coordinate function of $f-v$ vanishes: over the algebraically closed field the closed points are dense on each affine open and their vanishing ideal is the nilradical. Therefore $f$ factors scheme-theoretically through $v$. Pulling back along $G\to G\mathrm{/}B$ shows that all of $G$ fixes $v$, so $G={\mathrm{Stab}}_G(v)=B$ is unipotent. This proves the rank-zero equivalence.

If $G$ has semisimple rank zero, its smooth connected affine quotient $Q=G\mathrm{/}R(G)$ has rank zero and is unipotent by step 3.1 and [F1]. It is therefore solvable. Since $R(G)$ is solvable, [F4] makes $G$ solvable, so $G=R(G)$ by maximality of the radical. This conclusion does not require asserting that $Q$ is geometrically semisimple over an imperfect field.

If in addition $G$ is reductive, pass to the algebraic closure. The solvable group $G_{k^{\mathrm{a}}}$ decomposes as $(G_{k^{\mathrm{a}}}{)}_u⋊T$ by [F2]. Its smooth connected normal unipotent factor is trivial by reductivity, so $G_{k^{\mathrm{a}}}=T$. Hence $G$ is a torus by the definition of a torus as a group becoming a split torus over an algebraic closure. This proves the final assertion.